On Thursday, 23 February 2017 at 18:33:33 UTC, Meta wrote:
On Thursday, 23 February 2017 at 18:21:51 UTC, Meta wrote:
On Thursday, 23 February 2017 at 16:01:44 UTC, John Colvin
wrote:
Is there any way to get a reference/alias to the
instantiation of a template function that would be called,
given certain parameters? I.e. to get the result of whatever
template parameter inference (and overload resolution) has
occurred?
E.g. for some arbitrarily complex foo:
static assert(__traits(compiles, foo(3)));
alias fooWithInt = someMagic(foo(3));
so if foo was `void foo(T)(T t) {}` then `fooWithInt` would
be `foo!int`, but if it was `void foo(Q = float, T = long)(T
t)` then `fooWithInt` would be `foo!(float, int)`
I don't believe so, because foo(3) is a value (void), not a
type like foo!int would be. You can't get it back after you've
called the function. You would have to do something like:
alias fooWithInt = someMagic!foo(3);
Where someMagic constructs the alias to foo!int based on the
type of arguments passed, or something like that.
A quick and rough example I threw together. Annoyingly, I can't
figure out a way to tell the compiler that I want to printout
the symbol of fooWithInt, not call it without parens. The best
I can do is print out its type.
void foo(T)(T t) {}
void foo(Q = float, T = long)(T t) {}
alias Typeof(alias v) = typeof(v);
template getInstantiation(alias f, T...)
{
import std.meta;
alias getInstantiation = f!(staticMap!(Typeof, T));
}
alias fooWithInt = getInstantiation!(foo, 3);
alias fooWithLong = getInstantiation!(foo, 3L);
void main()
{
pragma(msg, typeof(fooWithInt));
pragma(msg, typeof(fooWithLong));
}
Unfortunately that only works by accident of my example. A
counterexample:
T foo(Q = float, T = short)(T t) { return t; }
alias Typeof(alias v) = typeof(v);
template getInstantiation(alias f, T...)
{
import std.meta;
alias getInstantiation = f!(staticMap!(Typeof, T));
}
static assert(is(typeof(foo(3)) == int)); // ok
static assert(is(typeof(getInstantiation!(foo, 3)(3)) == int));
// fails